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Proof step 3 from proof cons-to-addnum (previous step)

⊦ ⌜a '⌝ + σ to n + m = ⌜a '⌝ + (σ to (n - 1) + m), n = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m0] flags:

Additional information

Used simplifier rules: to-succ-cons: ⊦ ⌜a '⌝ + σ to n + 1 = a + (σ to n);%cons-append: ⊦ a + σ = ⌜a '⌝ + σ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%append-cancel-l: ⊦ ⌜x⌝ + σ0 = ⌜x⌝ + σ1 ↔ σ0 = σ1;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0